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Are there any other functions (not necessarily continuous) satisfying $f(xy)=f(x)+f(y)$ other than $f(x)=A \ln x$ and $f(x)=0$?

After a little thought I came to identify a function $$f : \mathbb{C} \to \mathbb{R}: z \mapsto \arg(z),$$ since $$\arg(z_1z_2)=\arg(z_1)+\arg(z_2).$$ Also $$f: A-\left\{0\right\} \to \mathbb{Z_0^+}$$ where $A$ is set of non-zero polynomials such that $$f(x)=\operatorname{Deg}(\text{polynomial}),$$ since $$\operatorname{Deg}(h(x)g(x))=\operatorname{Deg}(h(x))+\operatorname{Deg}(g(x)).$$

Are there other functions with this property?

Xander Henderson
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Ekaveera Gouribhatla
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    There is no global argument function that would satisfy this. – Arnaud Mortier Jul 07 '18 at 15:53
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    Let $F(x)$ be any function satisfying Cauchy's functional equation. Then the function $f(x)=F\left(\ln (x) \right)$ satisfies your functional equation. – lulu Jul 07 '18 at 15:59
  • Worth noting: if you add some sort of regularity condition you should be able to eliminate the pathological examples. Continuity at a single point is probably strong enough. – lulu Jul 07 '18 at 16:02
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    The answer depends on the domain you want $f$ to have. If $0$ is in your domain, then $f(x\cdot 0)=f(x)+f(0)$ implies $f(x)=0$ for all $x$. So for your $\arg$ example, you have to at least switch to domain $\Bbb C^\times$. Even after that, $0=\arg 1=\arg((-1)\cdot(-1))\ne \arg(-1)+\arg(-1)=2\pi$ – Hagen von Eitzen Jul 07 '18 at 16:05
  • Note that assuming continuity over $\mathbb R$ and positive $x,y$, the only function that satisfied this is $\ln$. See here.and here. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jul 07 '18 at 16:16
  • The $p$-adic valuation for primes $p$ has this property. – Bilbottom Jul 07 '18 at 16:50
  • Any discrete valued function possible? – Ekaveera Gouribhatla Jul 07 '18 at 17:15
  • If $0$ is in the domain then $f(x) = f(x) + f(0) - f(0) = f(0*x) - f(0)=f(0)-f(0) = 0$. Are we to assume the domain is closed under multiplication? I could defined $Domain \subset \mathbb R$ so that for any $x,y \in Domain$ that $xy \not \in Domain$ and define whatever I darn well want. And I can have some limited products. Example $f:{1,5,7,35,49}$, $f(1)=0; f(5)=\pi; f(7)=2;f(35)=\pi + 2; f(49) = 4$. will work. I could even close that under multiplication. $Domain={5^n7^m}$ $f(5^n7^m)=2n+m\pi$. Do we assume the domain includes multiplicative inverses? – fleablood Jul 07 '18 at 17:26
  • @TheSimpliFire: How do you exclude logarithms with other bases? – hmakholm left over Monica Jul 07 '18 at 17:39
  • @HenningMakholm other bases are just base $e$ times a constant, so they will work – ℋolo Jul 07 '18 at 18:12
  • @Holo: So I would think too. But TheSimpliFire is claiming that the natural logarithm specifically is the only solution. – hmakholm left over Monica Jul 07 '18 at 18:14
  • Let $f,g$ satisfy the condition, we can easily see that $Af+Bg$ also satisfy the condition: $H(xy)=Af(xy)+Bg(xy)=A(f(x)+f(y))+B(g(x)+g(y))=Af(x)+Bg(x)+Af(y)+Bg(y)=H(x)+H(y)$, also, like @lulu said, let $F$ be function that satisfy Cauchy's functional equation then $F\circ g$ satisfy the conditions. Also, for the positive integers it is called Completely additive function – ℋolo Jul 07 '18 at 18:37

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If you have a function $f$ such that $f(xy)=f(x)+f(y)$, then $g(x)=f(e^{x})$ satisfies $g(x+y)=g(x)+g(y)$. There are highly discontinuous functions $g$ that satisfy this identity, but they require the Axiom of Choice, which is a standard Axiom of Mathematics. For any such $g$, the function $h(x)=g(\ln x)$ is discontinuous on $(0,\infty)$ and satisfies the following $$ h(xy)=g(\ln(xy))=g(\ln x+\ln y)=g(\ln x)+g(\ln y)= h(x)+h(y),\;\; x,y > 0. $$

Disintegrating By Parts
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